2 + 5 + 10 + 17 + 26 + ………..
Solution:
Let Tn and Sn be the nth term and the sum of n terms respectively, of the given series.
Then, we have:
$ {{S}_{n}}~=\text{ }2\text{ }+\text{ }5\text{ }+\text{ }10\text{ }+\text{ }17\text{ }+\text{ }26\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}}~\ldots \text{ }\left( 1 \right) $
we can rewrite equation (1) as
$ {{S}_{n}}~=\text{ }2\text{ }+\text{ }5\text{ }+\text{ }10\text{ }+\text{ }17\text{ }+\text{ }26\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}}~\ldots \ldots ..\left( 2 \right) $
Upon subtracting (2) from (1) we get
$ {{S}_{n}}~=\text{ }2\text{ }+\text{ }5\text{ }+\text{ }10\text{ }+\text{ }17\text{ }+\text{ }26\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}} $
$ {{S}_{n}}~=\text{ }2\text{ }+\text{ }5\text{ }+\text{ }10\text{ }+\text{ }17\text{ }+\text{ }26\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}} $
$ 0\text{ }=\text{ }2\text{ }+\text{ }\left[ 3\text{ }+\text{ }5\text{ }+\text{ }7\text{ }+\text{ }9\text{ }+\text{ }\ldots \text{ }+\text{ }\left( {{T}_{n}}~\text{ }{{T}_{n-1}} \right) \right]-{{T}_{n}} $
The successive terms differ by 3, 5, 7, 9. As we can see that these differences are in A.P
We have:
Therefore, the sum of the series is n/6 (2n2 + 3n + 7)